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in triangle ABC , D is the midpoint of AB . Pt.E is on AC such that AE=2EC . BE intersects CD at pt.F

Prove that BE=4EF

6 years ago


Answers : (1)


Dear ajinkya

Let A be (0,0),  B(x1,y1),  C(x2,y2)

then  By section formula 



Let F(x3,y3)

let BF:FE=k:1    and DF:FC=m:1

then by section formula on line BE and CD we get

on BE........x3=(k*(2x2/3)+x1)/(k+1)  and y3= (k*(2y2/3)+y1)/(k+1)           ......(1)

On CD.......x3=(m*x2+x1/2)/(m+1)  and y3=(m*y2+y1/2)/(m+1)                        .........(2)

from (1) and (2)

(k*(2x2/3)+x1)/(k+1) =(m*x2+x1/2)/(m+1)        and     (k*(2y2/3)+y1)/(k+1) =(m*y2+y1/2)/(m+1) 

Compare to get coefficients of x1 and x2 or y1 and y2

we get


and 1/(k+1)=1/2(m+1)

solve these two eq. to get k=3;

So BF:FE=3:1



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Askiitians Experts

Sahil Arora
IIT Delhi
6 years ago

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